S_8(s) + 12O_2(g) rarr 8SO_3(g); (i)
Alternatively,
1/8S_8(s) + 3/2O_2(g) rarr SO_3(g) (ii)
Note that this gives the stoichiometric conversion of S(0) to S(VI). Industrially sulfur trioxide (and what do you think this is used for?) derives from SO_2(g).
i.e. 1/8S_8(s) +O_2(g) rarr SO_2(g); (iii)
then SO_2(g) + 1/2O_2(g) rarr SO_3(g) (iv)
In the third equation I used the 1/8 coefficient to cope with the molecularity of the sulfur ring (then I used a 1/2 coefficient in the fourth equation; I am certainly free to do this). I am not free to leave the equation unbalanced. Is it balanced? Don't trust my arithmetic. The fourth equation to give SO_3 uses (I think) a supported V_2O_5 catalyst of some sort. It must be an incredibly dirty and smelly process, yet it is undoubtedly vital to our civilization.
Note that SO_3(g) is often added to sulfuric acid to give H_2S_2O_7, pyrosulfuric acid.
